Halogenated compounds are particularly easy to identify by their mass spectra because chlorine and bromine occur naturally as mixtures of two abundant isotopes. Chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); Bromine occurs as 79Br (50.7%) and 81Br (49.3%); Boron compounds also stand out owing to the two isotopes 10B (19.9%) and 11B (80.1%). For the compound Bromomethane, CH3Br: At what masses do the molecular ions occur

See answer below

Explanation:

In this case, we know that the molecular formula of bromomethane is CH₃Br

To know the the molecular ions, we need to take in account both isotopes of the bromine, and calculate the molecular mass of bromomethane:

The molecular mass of carbon and hydrogen are:

C: 12 g/mol;   H: 1 g/mol

For Br -79:

CH₃Br = (1*12) + (3*1) + 79 = 94 g/mol

For Br-81:

CH₃Br = (1*12) + (3*1) + 81 = 96 g/mol

Hope this helps

the answer is b. dilution